Q. 7.37 - A First Course in Probability

Question

An urn contains$a$ white and $b$black balls. After a ball is drawn, it is returned to the urn if it is white; but if it is black, it is replaced by a white hall from another urn. Let $M_n$ denote the expected number of white balls in the urn after the foregoing operation has been repeated$n$ times.

1.  Derive the recursive equation $M_{n+1}=\left(1-\frac{1}{a+b}\right)M_n+1$
   
2. Use part (a) to prove that $M_n=a+b-b{\left(1-\frac{1}{a+b}\right)}^n$
   
3. What is the probability that the $(n+1)$ ball drawn is white?

Step-by-Step Solution

Answer

Use basic properties of conditional expectation to obtain the required expressions and values.

1. Applying expectation to both sides and using the relation, we get$M_{n+1}=\left(1-\frac{1}{a+b}\right)M_n$
   
2. The claimed expression hold for every$n\ge 0$
   
3. Applying the expectation to both sides, we have that $P\left(I_{n+1}=1\right)=E\left(I_{n+1}\right)=\frac{1}{a+b}{M}_n=\frac{1}{a+b}\left(a+b-b{\left(1-\frac{1}{a+b}\right)}^n\right)$
   

Step 1: Given Information (Part-a)

Given in the question that the define random variables $X_n$ which counts how many of white balls is in the urn after $n$ draws. The recursive equation  $M_{n+1}=\left(1-\frac{1}{a+b}\right)M_n+1$

Step 2: Find the Conditional Expectation (Part-a)

Let's find the conditional expectation of $X_{n+1}$given $X\_\left\{n\right\}.$ 

Consider what has been drawn in the $n+1$draw. If we have drawn a white ball (probability $X_n/(a+b)$), the expected value  $X_{n+1}$ is $X_n$ since we return back that white ball. If we have drawn a black ball (probability $1-X\_\left\{n\right\} /(a+b) ),$ we replace it with the white ball, so the expected value of $X_{n+1}$ is$X\_\left\{n\right\}+1.$ 

Hence

$E\left(X_{n+1}\mid X_n\right)=\frac{X_n}{a+b}\times X_n+\left(1-\frac{X_n}{a+b}\right)\times \left(X_n+1\right)$

$=\left(1-\frac{1}{a+b}\right)X_n$

Step 3: Final Answer (Part-a)

Applying expectation to both sides and using the relation $E\left(E\left(X_{n+1}\mid X_n\right)\right)=E\left(X_{n+1}\right)=M_{n+1}$

We get, $M_{n+1}=\left(1-\frac{1}{a+b}\right)M_n$ which has been claimed.

Step 4: Given Information (Part-b)

Given in the question to prove that  $M_n=a+b-b{\left(1-\frac{1}{a+b}\right)}^n$

Step 5: Prove The Equation (Part-b)

Let's prove it by induction. 

For $n=0$,

We have that$M_0=a+b-b=a$

which is true since we have $a$ white balls at the beginning. Suppose that the expression is true for$M_n$. Consider $M_{n+1}$. Using the relation from part (a), we have that

$M_{n+1}=\left(1-\frac{1}{a+b}\right)M_n$

$=\left(1-\frac{1}{a+b}\right)\left(a+b-b{\left(1-\frac{1}{a+b}\right)}^n\right)=a+b-b{\left(1-\frac{1}{a+b}\right)}^{n+1}$

Step 6: Final Answer (Part-b)

The claimed expression hold for every $n\ge 0$

Step 7: Given Information (Part-c)

Define indicator random variable $I_{n+1}$ that is equal to one if and only if the$n+1$drawn ball is white.

Let's find the conditional expectation of $I_{n+1}$ a given$X_n$.

We have that, $E\left(I_{n+1}\mid X_n\right)=\frac{X_n}{a+b}$

since if we know that there are $X_n$ white balls, we know that the probability that we draw white ball is $X_n/(a+b)$.

Step 8: Final Answer (Part-c)

Applying the expectation to both sides, we have that$P\left(I_{n+1}=1\right)=E\left(I_{n+1}\right)=\frac{1}{a+b}{M}_n=\frac{1}{a+b}\left(a+b-b{\left(1-\frac{1}{a+b}\right)}^n\right)$